You are misreading the graph, and setting up your equation incorrectly as well.
The skater starts off with an initial speed of 2.6 m/s^2 at a height 2 .5 m above a reference plane. For part a, at point A, the skater is 1.5 m above ground at some speed. The delta h is -1.0 m. But be consistent in setting up the energy equation. Either use initial KE plus initial PE = final KE plus final PE, where the initial PE is 2.5mg and the final PE is 1.5mg, or use the "delta" formula I referenced, where the change in PE is (1.5mg - 2.5 mg) = -1.0mg.

You are misreading the graph, and setting up your equation incorrectly as well.
The skater starts off with an initial speed of 2.6 m/s^2 at a height 2 .5 m above a reference plane. For part a, at point A, the skater is 1.5 m above ground at some speed. The delta h is -1.0 m. But be consistent in setting up the energy equation. Either use initial KE plus initial PE = final KE plus final PE, where the initial PE is 2.5mg and the final PE is 1.5mg, or use the "delta" formula I referenced, where the change in PE is (1.5mg - 2.5 mg) = -1.0mg.

The way i did it above is how my text book told me to do it. I can take a picture and post it of you would liek. Of an example. Because I am very lost now. So I can use hate equation I just did above ?

The way i did it above is how my text book told me to do it. I can take a picture and post it of you would liek. Of an example. Because I am very lost now. So I can use hate equation I just did above ?

Please use the first equation I gave you: 1/2m(2.6)^2 + mg(2.5) = 1/2mv^2 + mg(1.5). Solve for v at point A. Continue... Don't confuse h with delta h

No, start at the top again for initial condition (v = 2.6 and h_top = 2.5) and the final condition is at B (v = ? and h_B =0), which you sort of indicated the first time except you called it delta h instead of h_B, h_B is the height at B above the reference plane and is = to 0.